---
title: "Convergence; coefficients of ±1.5 square wave"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:square-wave-convergence-coefficients"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-convergence-coefficients
site: BCT Engineering
---

# Convergence; coefficients of ±1.5 square wave

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2080 Chaitra, 3+5 marks
- Video: [Finding Fourier coefficients for square wave](https://www.youtube.com/watch?v=WTD8jg8fXoE) — Khan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-convergence-coefficients
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Write down the condition of convergence of continuous time Fourier series. Find the Fourier series coefficient a_k for the continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2.

## Short answer

Dirichlet conditions: absolute integrability, finite extrema, finite discontinuities. For period T=2, ω0=π, integration gives a0=0, a_k = 3/(j kπ) for odd k, 0 for even k.

## Answer

#### Condition of convergence (Dirichlet conditions)
The Fourier series of a periodic $x(t)$ converges if, over one period:
1. $x(t)$ is absolutely integrable: $\int_T|x(t)|dt < \infty$.
2. $x(t)$ has a finite number of maxima and minima.
3. $x(t)$ has a finite number of finite discontinuities.

Then the series equals $x(t)$ wherever $x(t)$ is continuous and equals $\frac{1}{2}[x(t^-) + x(t^+)]$ at a jump. (Alternatively, finite energy in one period gives convergence in the mean-square sense.)

#### Fourier series coefficients
Period $T = 2$ s, so $\omega_0 = 2\pi/T = \pi$ rad/s.

$$
x(t) = \begin{cases} 1.5, & 0 \le t < 1 \\ -1.5, & 1 \le t < 2 \end{cases}
$$

**DC term:**

$$
a_0 = \frac{1}{2}\left[\int_0^1 1.5\,dt + \int_1^2(-1.5)\,dt\right] = \frac{1}{2}(1.5 - 1.5) = 0
$$

**For $k \ne 0$:**

$$
\begin{aligned}
a_k &= \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right] \\
&= \frac{1.5}{2}\left[\frac{1 - e^{-jk\pi}}{jk\pi} - \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi}\right]
\end{aligned}
$$

Using $e^{-j2k\pi} = 1$ and $e^{-jk\pi} = (-1)^k$:

$$
\begin{aligned}
a_k &= \frac{0.75}{jk\pi}\left[1 - (-1)^k - (-1)^k + 1\right] = \frac{0.75 \times 2\left[1 - (-1)^k\right]}{jk\pi} \\
&= \frac{1.5\left[1 - (-1)^k\right]}{jk\pi}
\end{aligned}
$$

So

$$
a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even (including } k = 0) \end{cases}
$$

| $k$ | $\pm1$ | $\pm2$ | $\pm3$ | $\pm5$ |
|---|---|---|---|---|
| $a_k$ | $\mp j0.955$ | 0 | $\mp j0.318$ | $\mp j0.191$ |
| $\lvert a_k\rvert$ | 0.955 | 0 | 0.318 | 0.191 |

Phase: $-90^\circ$ for positive odd $k$ and $+90^\circ$ for negative odd $k$ (the signal is odd, so the $a_k$ are purely imaginary).

Equivalent trigonometric form:

$$
x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]
$$

**Answer:** $a_0 = 0$; $a_k = \dfrac{3}{jk\pi}$ for odd $k$; $a_k = 0$ for even $k$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For x(t) = 1.5 on 0<=t<1 and -1.5 on 1<=t<2, what are T and w0?
   A. T = 1, w0 = 2 pi
   B. T = 2, w0 = pi
   C. T = 2, w0 = 2 pi
   D. T = 3, w0 = 2 pi/3
   Answer: B. One full cycle takes 2 s, so T = 2 and w0 = 2 pi/T = pi rad/s. (Key point: Fundamental frequency)

2. Why is a_0 = 0 for this signal?
   A. Because the signal is even
   B. Because the period is 2
   C. Because the positive and negative halves have equal areas that cancel
   D. Because the signal is discontinuous
   Answer: C. a_0 is the average: (1/2)(1.5 - 1.5) = 0. (Key point: DC coefficient)

3. What is a_k for even k and for odd k?
   A. a_k = 0 for even k; 3/(jk pi) for odd k
   B. a_k = 3/(jk pi) for all k
   C. a_k = 0 for odd k; 3/(jk pi) for even k
   D. a_k = 1.5 for odd k; 0 for even k
   Answer: A. a_k = 1.5[1 - (-1)^k]/(jk pi), which is zero for even k and 3/(jk pi) for odd k. (Key point: Even harmonics vanish)

4. At a jump discontinuity, a convergent Fourier series equals:
   A. The larger of the two one-sided limits
   B. Zero
   C. The left-hand limit only
   D. The average of the left and right limits
   Answer: D. At a jump the series converges to (1/2)[x(t-) + x(t+)]. (Key point: Convergence at a jump)

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